If n = 4p, where p is a prime number greater than 2, how many differen

Is one not a divisor?

It is but its' not even (how many different positive even divisors does n have, including n).

p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?

No matter which prime p is, 4p will have only four EVEN divisors: 2, 4, 2p, and 4p. Try to check it with any prime greater than 2.

Does this make sense?

4 has two even divisors >> 2 & 4

Any prime no p divisors will have 2p & 4p (from above)

So total = 4

n=4*p

And p(prime) > 2

n= 2*2*p

Positive even divisors 'n' can have including 'n':

2
2p
4
4p which is also equal to 'n'

Hence 4 positive even divisors.

Yes, your approach is correct.

@p=3, n = 4*3 = 12, Positive Even divisor of n = {2, 4, 6, 12} i.e. 4 Divisors
@p=5, n = 4*5 = 20, Positive Even divisor of n = {2, 4, 10, 20} i.e. 4 Divisors
@p=7, n = 4*7 = 28, Positive Even divisor of n = {2, 4, 14, 28} i.e. 4 Divisors
@p=11, n = 4*11 = 44, Positive Even divisor of n = {2, 4, 22, 44} i.e. 4 Divisors
@p=13, n = 4*13 = 52, Positive Even divisor of n = {2, 4, 26, 52} i.e. 4 Divisors
and so on...

Answer: option C

Hi All,

This question can be solved rather easily by TESTing VALUES:

We're told that N = 4P and that P is a prime number greater than 2. Let's TEST P = 3; so N = 12

The question now asks how many DIFFERENT positive EVEN divisors does 12 have, including 12?

12:
1,12
2,6
3,4

How many of these divisors are EVEN? 2, 4, 6, 12 …..4 even divisors.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi All,

This question can be solved rather easily by TESTing VALUES:

We're told that N = 4P and that P is a prime number greater than 2. Let's TEST P = 3; so N = 12

The question now asks how many DIFFERENT positive EVEN divisors does 12 have, including 12?

12:
1,12
2,6
3,4

How many of these divisors are EVEN? 2, 4, 6, and 12 …..that's a total of 4 even divisors.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

n = 4p
p prime no. \(> 2\)

We have to find the no. of even divisors which means even factors of n

P must be a odd no. because 2 is the only even prime no.

Let p be \(3\)

\(n= 4 * 3\)

= \(2^{2} * 3\)

Now , No of factors of P = \(2^{2+1} * 3^{1+1}\) = \(2^{3} * 3^{2}\)

= \(3 * 2 = 6\)

Even factors = Total factors - odd factors

To find Odd factors we take all the prime apart from 2. so here we are left with only 3

Odd factors = \(3^{1+1} = 3^{2}= 2\)

Total factors - Odd factors = Even factors
\(6-2 = 4\)

hence answer is 4

MATH REVOLUTION VIDEO SOLUTION:

This is an interesting question because we are immediately given the option to insert any prime number we wish for p. Since this is a problem-solving question, and there can only be one correct answer, we can select any value for p, as long as it is a prime number greater than 2. We always want to work with small numbers, so we should select 3 for p. Thus, we have:

n = 4 x 3

n = 12

Next we have to determine all the factors, or divisors, of P. Remember the term factor is synonymous with the term divisor.

1, 12, 6, 2, 4, 3

From this we see that we have 4 even divisors: 12, 6, 2, and 4.

If you are concerned that trying just one value of p might not substantiate the answer, try another value for p. Let’s say p = 5, so

n = 4 x 5

n = 20

The divisors of 20 are: 1, 20, 2, 10, 4, 5. Of these, 4 are even: 20, 2, 10 and 4. As we can see, again we have 4 even divisors.

No matter what the value of p, as long as it is a prime number greater than 2, n will always have 4 even divisors.

The answer is C.

We can solve a lot of Integer Properties questions by testing a value

If p is prime, let's let p = 3
So, n = 4p = (4)(3) = 12

The positive EVEN divisors of 12 are: 2, 4, 6 and 12
So, there are FOUR even divisors of n.

Answer: C

Cheers,
Brent

Bunuel in case of n=4pq
Should not there be condition that p and Q must be " distinct" prime numbers and both greater than 2.

Number of even divisors.

Since p is prime>2, p on its own is not an even divisor.

So the possibilities are 4p, 2p, 4, 2. Answer is C

2, 4, 2p & 4p
Picking numbers drains brain energy very fast I try to avoid it as much as possible